第二个PHP代码片段不工作 - 2nd snippet of PHP code not working

2nd snippet of PHP code not working

本文关键字：工作片段代码 PHP 第二个 | 更新日期: 2023-09-27

我有一个PHP文件，其中有两组PHP代码。第一个还行，第二个就不行了。该文件完美地呈现了第一组php代码，然后呈现了一个表单，然后，它没有运行第二组php代码，而是将其打印出来，下面是代码:

<!DOCTYPE html>
<head>
    <title>Rate your favorite movie</title>
    <link rel='stylesheet' type='text/css' href='//maxcdn.bootstrapcdn.com/bootstrap/3.2.0/css/bootstrap.min.css'>
</head>
<html>
<body>
    <div class="container">
        <div class="row">
            <div class="col-md-2"> </div>   
            <div class="col-md-5" style="background:#2b1b17;color:white">
                <?php  //  FIRST SET OF PHP CODE
                    function get_movie_information($name) {
                        $url = "http://www.omdbapi.com/?t=".urlencode($name); 
                        // send request 
                        $curl = curl_init();
                        curl_setopt($curl, CURLOPT_URL, $url);
                        curl_setopt($curl, CURLOPT_FOLLOWLOCATION, 1);
                        curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1);
                        $curlData = curl_exec($curl);
                        curl_close($curl);
                        return json_decode($curlData, true);
                    }
                    $arr = get_movie_information($_POST["name"]);
                    echo "<br>"; 
                    $poster = $arr['Poster'];
                    echo '<img src="' . $poster . '">';
                    echo "<br>";
                    echo $arr['Title'];
                    echo "<br>";
                    echo $arr['Year'];
                    echo "<br>";
                    echo $arr['Rated'];
                    echo "<br>";
                    echo $arr['Genre'];
                    echo "<br>";
                    echo $arr['Plot'];
                    echo "<br>";
                ?>
                <form Id='Form' Method='post' Action='index.php'>
                    <input type="submit" value="Add Movie">
                </form>
                <?   // SECOND SET OF PHP CODE
                    $hostname='127.0.0.1';
                    $username='root';
                    $password='password';
                    If(isset($_POST['submit'])){
                        try {
                            $db = new PDO("mysql:host=$hostname;dbname=baboon;charset=utf8", $username, $password);
                            $db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
                            $stmt = $db->prepare("INSERT INTO movies (title, year, rated, plot) VALUES (:title, :year, :rated, :plot)");
                            // old code that didnt work
                            //  $stmt->bindParam(':title', $title); 
                            //  $stmt->bindParam(':year', $year);
                            //  $stmt->bindParam(':rated', $rated);
                            //  $stmt->bindParam(':plot', $plot);
                            $stmt->execute(array(
                                $title = $arr['Title'];
                                $year =  $arr['Year'];
                                $rated = $arr['Rated'];
                                $plot =  $arr['Plot'];
                            ));
                        } catch(PDOException $e) {
                            echo "Error: " . $e->getMessage();
                        }
                        $db = null;
                    }
                ?>
            </div>
        </div>
    </div>

    <script src="//maxcdn.bootstrapcdn.com/bootstrap/3.2.0/js/bootstrap.min.js"></script>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
</body>
</html>

$db->setAttribute之后的内容都打印出来了

我在审查你的代码时注意到的一些问题:

您的<head></head>标签在<body></body>标签之外。
你的第二个php标签被标记为<?而不是<?php像你的第一个php标签。
你在第二个php代码中的if有一个大写的'I'。